$\lim_{x\to\infty}\dfrac{4x^2-7x}{(\ln(x))^2}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $4$ (Choice C) C $8$ (Choice D) D $\infty$
$\lim_{x\to\infty} 4x^2-7x=\infty$ and $\lim_{x\to\infty} (\ln(x))^2=\infty$, so $\lim_{x\to\infty}\dfrac{4x^2-7x}{(\ln(x))^2}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{4x^2-7x}{(\ln(x))^2} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[4x^2-7x\right]}{\dfrac{d}{dx}[(\ln(x))^2]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{8x-7}{2\ln(x)\cdot\dfrac{1}{x}} \\\\ &=\lim_{x\to\infty}\dfrac{8x^2-7x}{2\ln(x)} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[8x^2-7x\right]}{\dfrac{d}{dx}[2\ln(x)]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{16x-7}{\left(\dfrac{2}{x}\right)} \\\\ &=\lim_{x\to\infty}\dfrac{16x^2-7x}{2} \\\\ &=\infty \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{\infty}{\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[8x^2-7x\right]}{\dfrac{d}{dx}[2\ln(x)]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{4x^2-7x}{(\ln(x))^2}=\infty$.